Consider the functions, [math]f_1\left(x,y\right)=\left(x,\frac{1}{y}\right)[/math] and [math]f_3\left(x,y\right)=\left(x+2y,y\right)[/math]. [br][br]We can show that [math]f_1[/math] is not an onto function because there are points in the plane that cannot be obtained as outputs of [math]f_1[/math]. Counterexample: The point (1,0) exists in the target but there does not exist a corresponding (x,y) in the domain. That is, there is no point (x,y) such that f(x,y)=(1,0). [br]We can show that [math]f_3[/math] is an onto function. Let [math]f_3\left(x,y\right)=\left(a,b\right)[/math]. Then [math]a=x+2y[/math] and [math]b=y[/math]. Both a and b are defined for all values of x and y. Thus, for any (a,b) in the target, there exists an (x,y) in the domain such that f(x,y)=(a,b). [br][br]We can easily show that [math]f_1[/math] is one-to-one. Consider the element [math]Y\in f_1[/math] such that [math]Y=\left(x_1,\frac{1}{y_1}\right)[/math]. We see that the x-coordinate always maps to itself, thus there is a one-to-one relationship for the x-coordinates. By rearranging, we see that [math]y_1=\frac{1}{y}[/math]. Taking the reciprocal of any real number gives another unique real number and thus the mapping between the y-coordinates is also one-to-one. That is, [math]f_1[/math] is one-to-one. [br]We can easily that [math]f_3[/math] is also one-to-one. Consider the element [math]Y\in f_3[/math] such that [math]Y=\left(x_1+2y_1,y_1\right)[/math], The y-coordinate always maps to itself, thus there is a one-to-one relationship for the y-coordinates. By rearranging, we see that [math]x_1=x-2y_1[/math]. This equation gives a unique real. Thus the mapping between the x-coordinates is one-to-one. Therefore, [math]f_3[/math] is one-to-one. [br]We can see that neither function is distance preserving. Consider the points A=(a,b) and B=(c,d). Then [math]d\left(A,B\right)=\sqrt{\left(a-c\right)^2+\left(b-d\right)^2}[/math]. For [math]f_1[/math], [math]d\left(f_1\left(A\right),f_1\left(B\right)\right)=\sqrt{\left(a-c\right)^2+\left(\frac{1}{b}-\frac{1}{d}\right)^2}\ne d\left(A,B\right)[/math]. [br]For [math]f_3[/math], [math]d\left(f_3\left(A\right),f_3\left(B\right)\right)=\sqrt{\left(a+2b-c-2d\right)^2+\left(b-d\right)^2}\ne d\left(A,B\right)[/math].